TITLE: Proving a trig infinite sum using integration
QUESTION [15 upvotes]: How can I prove the following using integration and elementary functions?
Prove that:
$$\sum_{n=1}^{\infty} \frac{\sin(n\theta)}{n} = \frac{\pi}{2} - \frac{\theta}{2}$$
$0 <  \theta <  2\pi$

REPLY [4 votes]: In this answer, I show

$$
\begin{align}
\sum_{k=1}^\infty\frac{\sin(2kx)}{k}
&=\sum_{k=1}^\infty\frac{e^{i2kx}-e^{-i2kx}}{2ik}\\
&=\frac1{2i}\left(-\log(1-e^{i2x})+\log(1-e^{-i2x})\right)\\
&=\frac1{2i}\log(-e^{-i2x})\\[4pt]
&=\frac\pi2-x\quad\text{for }x\in\left(0,\pi\right)
\end{align}
$$

which is, in essence, milind's answer. However, the question asks about integration. This sounds as if the question is asking to find the Fourier Series of $f(\theta)=\frac\pi2-\frac\theta2$. First, note that $f(\theta)$ is odd; that is,
$$
\begin{align}
f(2\pi-\theta)
&=\frac\pi2-\frac{2\pi-\theta}2\\
&=\frac\theta2-\frac\pi2\\
&=-\left(\frac\pi2-\frac\theta2\right)\\[6pt]
&=-f(\theta)\tag{1}
\end{align}
$$
Equation $(1)$ implies that
$$
\begin{align}
\color{#00A000}{\int_0^{2\pi}f(\theta)\cos(n\theta)\,\mathrm{d}\theta}
&=\int_0^{2\pi}f(2\pi-\theta)\cos(n 2\pi-n\theta)\,\mathrm{d}\theta\\
&=-\color{#00A000}{\int_0^{2\pi}f(\theta)\cos(n\theta)\,\mathrm{d}\theta}\\[6pt]
&=0\tag{2}
\end{align}
$$
because $\color{#00A000}{x}=-\color{#00A000}{x}\implies\color{#00A000}{x}=0$.
Now the question is
$$
\begin{align}
\int_0^{2\pi}f(\theta)\sin(n\theta)\,\mathrm{d}\theta
&=\int_0^{2\pi}\left(\frac\pi2-\frac\theta2\right)\sin(n\theta)\,\mathrm{d}\theta\\
&=-\frac1n\int_0^{2\pi}\left(\frac\pi2-\frac\theta2\right)\,\mathrm{d}\cos(n\theta)\\
&=\left.-\frac1n\left(\frac\pi2-\frac\theta2\right)\cos(n\theta)\right]_0^{2\pi}\\
&\hphantom{=\,}+\frac1n\int_0^{2\pi}\cos(n\theta)\,\mathrm{d}\left(\frac\pi2-\frac\theta2\right)\\[4pt]
&=\frac\pi{n}\tag{3}
\end{align}
$$
$(3)$ says that the Fourier series for $f(\theta)$ on $(0,2\pi)$ is
$$
f(\theta)=\sum_{n=1}^\infty\frac{\sin(n\theta)}{n}\tag{4}
$$